伍德里奇计量经济学第六版答案Chapter 2

CHAPTER 2

TEACHING NOTES

This is the chapter where I expect students to follow most, if not all, of the algebraic derivations. In class I like to derive at least the unbiasedness of the OLS slope coefficient, and usually I derive the variance. At a minimum, I talk about the factors affecting the variance. To simplify the notation, after I emphasize the assumptions in the population model, and assume random sampling, I just condition on the values of the explanatory variables in the sample. Technically, this is justified by random sampling because, for example, E(u i|x1,x2,…,x n) = E(u i|x i) by independent sampling. I find that students are able to focus on the key assumption SLR.4 and subsequently take my word about how conditioning on the independent variables in the sample is harmless. (If you prefer, the appendix to Chapter 3 does the conditioning argument carefully.) Because statistical inference is no more difficult in multiple regression than in simple regression, I postpone inference until Chapter 4. (This reduces redundancy and allows you to focus on the interpretive differences between simple and multiple regression.)

You might notice how, compared with most other texts, I use relatively few assumptions to derive the unbiasedness of the OLS slope estimator, followed by the formula for its variance. This is because I do not introduce redundant or unnecessary assumptions. For example, once SLR.4 is assumed, nothing further about the relationship between u and x is needed to obtain the unbiasedness of OLS under random sampling.

Incidentally, one of the uncomfortable facts about finite-sample analysis is that there is a difference between an estimator that is unbiased conditional on the outcome of the covariates and one that is unconditionally unbiased. If the distribution of the is such that they can all equal the same value with positive probability – as is the case with discreteness in the distribution –then the unconditional expectation does not really exist. Or, if it is made to exist then the estimator is not unbiased. I do not try to explain these subtleties in an introductory course, but I have had instructors ask me about the difference.

SOLUTIONS TO PROBLEMS

2.1 (i) Income, age, and family background (such as number of siblings) are just a few

possibilities. It seems that each of these could be correlated with years of education. (Income and education are probably positively correlated; age and education may be negatively correlated because women in more recent cohorts have, on average, more education; and number of siblings and education are probably negatively correlated.)

(ii) Not if the factors we listed in part (i) are correlated with educ . Because we would like to hold these factors fixed, they are part of the error term. But if u is correlated with educ then E(u|educ ) ≠ 0, and so SLR.4 fails.

2.2 In the equation y = β0 + β1x + u , add and subtract α0 from the right hand side to get y = (α0 + β0) + β1x + (u - α0). Call the new error e = u - α0, so that E(e ) = 0. The new intercept is α0 + β0, but the slope is still β1.

2.3 (i) Let y i = GPA i , x i = ACT i , and n = 8. Then x = 25.875, y =

3.2125, 1n

i =∑(x i – x )(y i – y ) =

5.8125, and 1

n

i =∑(x i – x )2 = 56.875. From equation (2.9), we obtain the slope as 1

?β= 5.8125/56.875 ≈ .1022, rounded to four places after the decimal. From (2.17), 0?β = y – 1

?βx ≈ 3.2125 – (.1022)25.875 ≈ .5681. So we can write

GPA = .5681 + .1022 ACT

n = 8.

The intercept does not have a useful interpretation because ACT is not close to zero for the population of interest. If ACT is 5 points higher, GPA increases by .1022(5) = .511.

(ii) The fitted values and residuals — rounded to four decimal places — are given along with the observation number i and GPA in the following table:

You can verify that the residuals, as reported in the table, sum to -.0002, which is pretty close to zero given the inherent rounding error.

(iii) When ACT = 20, GPA = .5681 + .1022(20) ≈ 2.61.

(iv) The sum of squared residuals,21

?n

i i u

=∑, is about .4347 (rounded to four decimal places), and the total sum of squares,1

n

i =∑(y i – y )2, is about 1.0288. So the R -squared from the regression

is

R 2 = 1 – SSR/SST ≈ 1 – (.4347/1.0288) ≈ .577.

Therefore, about 57.7% of the variation in GPA is explained by ACT in this small sample of students.

2.4 (i) When cigs = 0, predicted birth weight is 119.77 ounces. When cigs = 20, bwght = 109.49. This is about an 8.6% drop.

(ii) Not necessarily. There are many other factors that can affect birth weight, particularly overall health of the mother and quality of prenatal care. These could be correlated with

cigarette smoking during birth. Also, something such as caffeine consumption can affect birth weight, and might also be correlated with cigarette smoking.

(iii) If we want a predicted bwght of 125, then cigs = (125 – 119.77)/( –.524) ≈–10.18, or about –10 cigarettes! This is nonsense, of course, and it shows what happens when we are trying to predict something as complicated as birth weight with only a single explanatory variable. The largest predicted birth weight is necessarily 119.77. Yet almost 700 of the births in the sample had a birth weight higher than 119.77.

(iv) 1,176 out of 1,388 women did not smoke while pregnant, or about 84.7%. Because we are using only cigs to explain birth weight, we have only one predicted birth weight at cigs = 0. The predicted birth weight is necessarily roughly in the middle of the observed birth weights at cigs = 0, and so we will under predict high birth rates.

2.5 (i) The intercept implies that when inc = 0, cons is predicted to be negative $124.84. This, of course, cannot be true, and reflects that fact that this consumption function might be a poor predictor of consumption at very low-income levels. On the other hand, on an annual basis, $124.84 is not so far from zero.

(ii) Just plug 30,000 into the equation: cons = –124.84 + .853(30,000) = 25,465.16 dollars.

(iii) The MPC and the APC are shown in the following graph. Even though the intercept is negative, the smallest APC in the sample is positive. The graph starts at an annual income level of $1,000 (in 1970 dollars).

increases housing prices.

(ii) If the city chose to locate the incinerator in an area away from more expensive neighborhoods, then log(dist) is positively correlated with housing quality. This would violate SLR.4, and OLS estimation is biased.

(iii) Size of the house, number of bathrooms, size of the lot, age of the home, and quality of the neighborhood (including school quality), are just a handful of factors. As mentioned in part (ii), these could certainly be correlated with dist [and log(dist )].

2.7 (i) When we condition on inc

becomes a constant. So E(u |inc

?e |inc

) = ?E(e |inc

?0 because E(e |inc ) = E(e ) = 0.

(ii) Again, when we condition on inc

becomes a constant. So Var(u |inc

?e |inc

2Var(e |inc ) = 2e σinc because Var(e |inc ) = 2e σ.

(iii) Families with low incomes do not have much discretion about spending; typically, a low-income family must spend on food, clothing, housing, and other necessities. Higher income people have more discretion, and some might choose more consumption while others more saving. This discretion suggests wider variability in saving among higher income families.

2.8 (i) From equation (2.66),

1β = 1n i i i x y =?? ???∑ / 21n i i x =?? ???

∑.

Plugging in y i = β0 + β1x i + u i gives

1β = 011()n i i i i x x u ββ=??++ ???∑/ 21n i i x =?? ???

∑.

After standard algebra, the numerator can be written as

2

0111

1

i

n n n

i i i i i i x x x u ββ===++∑∑∑.

Putting this over the denominator shows we can write 1β as

1β = β01n i i x =?? ???∑/ 21n i i x =?? ???

∑ + β1 + 1n i i i x u =?? ???∑/ 21n i i x =??

???∑.

Conditional on the x i , we have

E(1β) = β01n i i x =?? ???∑/ 21n i i x =??

???

∑ + β1

because E(u i ) = 0 for all i . Therefore, the bias in 1β is given by the first term in this equation. This bias is obviously zero when β0 = 0. It is also zero when 1n

i i x =∑ = 0, which is the same as

x = 0. In the latter case, regression through the origin is identical to regression with an intercept. (ii) From the last expression for 1βin part (i) we have, conditional on the x i ,

Var(1β) = 221n i i x -=?? ???∑Var 1n i i i x u =?? ???∑ = 2

21n i i x -=?? ???∑21Var()n i i i x u =??

???∑

= 2

21n i i x -=?? ???∑221n i i x σ=?? ???∑ = 2σ/ 21n i i x =?? ???

∑.

(iii) From (2.57), Var(1?β) = σ2/21()n i i x x =??- ???

∑. From the hint, 21n i i x =∑ ≥ 21()n

i i x x =-∑, and so

Var(1β) ≤ Var(1

?β). A more direct way to see this is to write 21

()n

i i x x =-∑ = 221

()n

i i x n x =-∑, which is less than 21

n

i i x =∑ unless x = 0.

(iv) For a given sample size, the bias in 1β increases as x increases (holding the sum of the

2i

x fixed). But as x increases, the variance of 1?β

increases relative to Var(1β). The bias in 1

β is also small when 0β is small. Therefore, whether we prefer 1β or 1?β on a mean squared error basis depends on the sizes of 0β, x , and n (in addition to the size of 21n

i i x =∑).

2.9 (i) We follow the hint, noting that 1c y = 1c y (the sample average of 1i c y is c 1 times the sample average of y i ) and 2c x = 2c x . When we regress c 1y i on c 2x i (including an intercept) we use equation (2.19) to obtain the slope:

22

11

121

1

12

22

22

211

1

1112

2

2

1

()()()()

()

()()()?.()

n n

i i

i

i

i i n

n

i

i i i n

i

i

i n

i

i c x c x c y c y c c x x y y c x c x c

x x x x y y c c c c x x ββ======----=

=

----=?=

-∑∑∑∑∑∑

From (2.17), we obtain the intercept as 0β = (c 1y ) – 1β(c 2x ) = (c 1y ) – [(c 1/c 2)1

?β](c 2x ) = c 1(y – 1

?β

x ) = c 10

?β) because the intercept from regressing y i on x i is (y – 1

?βx ).

(ii) We use the same approach from part (i) along with the fact that 1()c y + = c 1 + y and

2()c x + = c 2 + x . Therefore, 11()()i c y c y +-+ = (c 1 + y i ) – (c 1 + y ) = y i – y and (c 2 + x i ) – 2()c x + = x i – x . So c 1 and c 2 entirely drop out of the slope formula for the regression of (c 1 + y i )

on (c 2 + x i ), and 1

β = 1

?β

. The intercept is 0

β = 1

()c y + – 1

β2

()c x + = (c 1 + y ) – 1

?β(c 2 + x ) = (1?y x β-) + c 1 – c 21?β = 0?β + c 1 – c 21

?β, which is what we wanted to show.

(iii) We can simply apply part (ii) because 11log()log()log()i i c y c y =+. In other words, replace c 1 with log(c 1), y i with log(y i ), and set c 2 = 0.

(iv) Again, we can apply part (ii) with c 1 = 0 and replacing c 2 with log(c 2) and x i with log(x i ). If 01垐 and ββ are the original intercept and slope, then 11?ββ= and 0021垐log()c βββ=-.

2.10 (i) This derivation is essentially done in equation (2.52), once (1/SST )x is brought inside the summation (which is valid because SST x does not depend on i ). Then, just define

/SST i i x w d =.

(ii) Because 111垐Cov(,)E[()] ,u u βββ=- we show that the latter is zero. But, from part (i),

()

1111?E[()] =E E().n

n i i i i i i u wu u w u u ββ==??-=????

∑∑ Because the i u are pairwise uncorrelated (they are independent), 22E()E(/)/i i u u u n n σ== (because E()0, i h u u i h =≠). Therefore,

(iii) The formula for the OLS intercept is 0

垐y x ββ=- and, plugging in 01y x u ββ=++ gives 0

1

1

1

1

垐?()().x u x u x β

ββββββ=++-=+--

(iv) Because 1

? and u β are uncorrelated, 222222201垐Var()Var()Var()/(/SST )//SST x x u x n x n x ββσσσσ=+=+=+,

which is what we wanted to show.

(v) Using the hint and substitution gives ()220?Var()[SST /]/SST x x

n x βσ=+ ()()

21222212

11/SST /SST .n

n

i x i x i i n x x x n x σσ--==??=-+=????∑∑

22111E()(/)(/)0.n n n

i i i i i i i w u u w n n w σσ======∑∑∑

2.11 (i) We would want to randomly assign the number of hours in the preparation course so that hours is independent of other factors that affect performance on the SAT. Then, we would collect information on SAT score for each student in the experiment, yielding a data set

{(,):1,...,}i i sat hours i n =, where n is the number of students we can afford to have in the study. From equation (2.7), we should try to get as much variation in i hours as is feasible.

(ii) Here are three factors: innate ability, family income, and general health on the day of the exam. If we think students with higher native intelligence think they do not need to prepare for the SAT, then ability and hours will be negatively correlated. Family income would probably be positively correlated with hours , because higher income families can more easily afford

preparation courses. Ruling out chronic health problems, health on the day of the exam should be roughly uncorrelated with hours spent in a preparation course.

(iii) If preparation courses are effective,1β should be positive: other factors equal, an increase in hours should increase sat .

(iv) The intercept, 0β, has a useful interpretation in this example: because E(u ) = 0, 0β is the average SAT score for students in the population with hours = 0.

2.12 (i) I will show the result without using calculus. Let be the sample average of the and write

2

2

1

1

2

2

0011

1

22

0011

22

01

()[()()]

()2()()()()2()()()()()n n

i

i

i i n

n

n

i i i i i n

n

i i i i n

i i y b y y y b y y y y y b y b y y y b y y n y b y y n y b ========-=-+-=-+--+-=-+--+-=-+-∑∑∑∑∑∑∑∑

where we use the fact (see Appendix A) that 1

()0n

i i y y =-=∑ always. The first term does not

depend on 0b and the second term,20()n y b -, which is nonnegative, is clearly minimized when 0b y =.

(ii) If we define i i u y y =-then 1

1

()n

n

i i i i u y y ===-∑∑and we already used the fact that this sum

is zero in the proof in part (i).

SOLUTIONS TO COMPUTER EXERCISES

C2.1 (i) The average prate is about 87.36 and the average mrate is about .732.

(ii) The estimated equation is

prate= 83.05 + 5.86 mrate

n = 1,534, R2 = .075.

(iii) The intercept implies that, even if mrate = 0, the predicted participation rate is 83.05 percent. The coefficient on mrate implies that a one-dollar increase in the match rate – a fairly large increase – is estimated to increase prate by 5.86 percentage points. This assumes, of course, that this change prate is possible (if, say, prate is already at 98, this interpretation makes no sense).

(iv) If we plug mrate = 3.5 into the equation we get ?

prate= 83.05 + 5.86(3.5) = 103.59. This is impossible, as we can have at most a 100 percent participation rate. This illustrates that, especially when dependent variables are bounded, a simple regression model can give strange predictions for extreme values of the independent variable. (In the sample of 1,534 firms, only 34 have mrate≥ 3.5.)

(v) mrate explains about 7.5% of the variation in prate. This is not much, and suggests that many other factors influence 401(k) plan participation rates.

C2.2 (i) Average salary is about 865.864, which means $865,864 because salary is in thousands of dollars. Average ceoten is about 7.95.

(ii) There are five CEOs with ceoten = 0. The longest tenure is 37 years.

(iii) The estimated equation is

salary= 6.51 + .0097 ceoten

log()

n = 177, R2 = .013.

We obtain the approximate percentage change in salary given ?ceoten = 1 by multiplying the coefficient on ceoten by 100, 100(.0097) = .97%. Therefore, one more year as CEO is predicted to increase salary by almost 1%.

C2.3 (i) The estimated equation is

sleep= 3,586.4 – .151 totwrk

n = 706, R2 = .103.

The intercept implies that the estimated amount of sleep per week for someone who does not work is 3,586.4 minutes, or about 59.77 hours. This comes to about 8.5 hours per night.

(ii) If someone works two more hours per week then ?totwrk = 120 (because totwrk is measured in minutes), and so sleep ?= –.151(120) = –18.12 minutes. This is only a few minutes a night. If someone were to work one more hour on each of five working days, sleep ?= –.151(300) = –45.3 minutes, or about five minutes a night.

C2.4 (i) Average salary is about $957.95 and average IQ is about 101.28. The sample standard deviation of IQ is about 15.05, which is pretty close to the population value of 15.

(ii) This calls for a level-level model: wage = 116.99 + 8.30 IQ

n = 935, R 2 = .096.

An increase in IQ of 15 increases predicted monthly salary by 8.30(15) = $124.50 (in 1980 dollars). IQ score does not even explain 10% of the variation in wage .

(iii) This calls for a log-level model:

log()wage = 5.89 + .0088 IQ

n = 935, R 2 = .099.

If ?IQ = 15 then log()wage ? = .0088(15) = .132, which is the (approximate) proportionate change in predicted wage. The percentage increase is therefore approximately 13.2.

C2.5 (i) The constant elasticity model is a log-log model:

log(rd ) = 0β + 1βlog(sales ) + u ,

where 1β is the elasticity of rd with respect to sales .

(ii) The estimated equation is log()rd = –4.105 + 1.076 log(sales )

n = 32, R 2 = .910.

The estimated elasticity of rd with respect to sales is 1.076, which is just above one. A one percent increase in sales is estimated to increase rd by about 1.08%.

C2.6 (i) It seems plausible that another dollar of spending has a larger effect for low-spending schools than for high-spending schools. At low-spending schools, more money can go toward purchasing more books, computers, and for hiring better qualified teachers. At high levels of spending, we would expend little, if any, effect because the high-spending schools already have high-quality teachers, nice facilities, plenty of books, and so on.

(ii) If we take changes, as usual, we obtain

1110log()(/100)(%),math expend expend ββ?=?≈?

just as in the second row of Table 2.3. So, if %10,expend ?=110/10.math β?=

(iii) The regression results are

2

1069.34 11.16 log()

408, .0297

math expend n R =-+==

(iv) If expend increases by 10 percent, 10math increases by about 1.1 percentage points. This is not a huge effect, but it is not trivial for low-spending schools, where a 10 percent increase in spending might be a fairly small dollar amount.

(v) In this data set, the largest value of math10 is 66.7, which is not especially close to 100. In fact, the largest fitted values is only about 30.2.

C2.7 (i) The average gift is about 7.44 Dutch guilders. Out of 4,268 respondents, 2,561 did not give a gift, or about 60 percent.

(ii) The average mailings per year is about 2.05. The minimum value is .25 (which presumably means that someone has been on the mailing list for at least four years) and the maximum value is 3.5.

(iii) The estimated equation is

2

2.01 2.65 4,268, .0138

gift mailsyear

n R =+==

(iv) The slope coefficient from part (iii) means that each mailing per year is associated with – perhaps even “causes” – an estimated 2.65 additional guilders, on average. Therefore, if each mailing costs one guilder, the expected profit from each mailing is estimated to be 1.65 guilders. This is only the average, however. Some mailings generate no contributions, or a contribution less than the mailing cost; other mailings generated much more than the mailing cost.

(v) Because the smallest mailsyear in the sample is .25, the smallest predicted value of gifts is 2.01 + 2.65(.25) ≈ 2.67. Even if we look at the overall population, where some people have received no mailings, the smallest predicted value is about two. So, with this estimated equation, we never predict zero charitable gifts.

C2.8 There is no “correct” answer to this question because all answers depend on how the

random outcomes are generated. I used Stata 11 and, before generating the outcomes on the i x , I set the seed to the value 123. I reset the seed to 123 to generate the outcomes on the i u . Specifically, to answer parts (i) through (v), I used the sequence of commands

set obs 500 set seed 123

gen x = 10*runiform() sum x

set seed 123

gen u = 6*rnormal() sum u

gen y = 1 + 2*x + u reg y x

predict uh, resid gen x_uh = x*uh sum uh x_uh gen x_u = x*u sum u x_u

(i) The sample mean of the i x is about 4.912 with a sample standard deviation of about 2.874.

(ii) The sample average of the i u is about .221, which is pretty far from zero. We do not get zero because this is just a sample of 500 from a population with a zero mean. The current sample is “unlucky” in the sense that the sample average is far from the population average. The sample standard deviation is about 5.768, which is nontrivially below 6, the population value.

(iii) After generating the data on i y and running the regression, I get, rounding to three decimal places,

0? 1.862β= and 1

? 1.870β= The population values are 1 and 2, respectively. Thus, the estimated intercept based on this sample of data is well above the population value. The estimated slope is somewhat below the population value, 2. When we sample from a population our estimates contain sampling error; that is why the estimates differ from the population values.

(iv) When I use the command sum uh x_uh and multiply by 500 I get, using scientific notation, sums equal to 4.181e-06 and .00003776, respectively. These are zero for practical purposes, and differ from zero only due to rounding inherent in the machine imprecision (which is unimportant).

(v) We already computed the sample average of the i u in part (ii). When we multiply by 500 the sample average is about 110.74. The sum of i i x u is about 6.46. Neither is close to zero, and nothing says they should be particularly close.

(vi) For this part I set the seed to 789. The sample average and standard deviation of the i x are about 5.030 and 2.913; those for the i u are about .077- and 5.979. When I generated the i y and run the regression I get

0?.701β= and 1

? 2.044β= These are different from those in part (iii) because they are obtained from a different random sample. Here, for both the intercept and slope, we get estimates that are much closer to the population values. Of course, in practice we would never know that.

计量经济学习题及答案

第一章绪论 一、填空题： 1．计量经济学是以揭示经济活动中客观存在的__________为内容的分支学科，挪威经济学家弗里希，将计量经济学定义为__________、__________、__________三者的结合。 2．数理经济模型揭示经济活动中各个因素之间的__________关系，用__________性的数学方程加以描述，计量经济模型揭示经济活动中各因素之间__________的关系，用__________性的数学方程加以描述。 3．经济数学模型是用__________描述经济活动。 4．计量经济学根据研究对象和内容侧重面不同，可以分为__________计量经济学和__________计量经济学。 5．计量经济学模型包括__________和__________两大类。 6．建模过程中理论模型的设计主要包括三部分工作，即__________、____________________、____________________。 7．确定理论模型中所包含的变量，主要指确定__________。 8．可以作为解释变量的几类变量有__________变量、__________变量、__________变量和__________变量。 9．选择模型数学形式的主要依据是__________。 10．研究经济问题时，一般要处理三种类型的数据：__________数据、__________数据和__________数据。 11．样本数据的质量包括四个方面__________、__________、__________、__________。 12．模型参数的估计包括__________、__________和软件的应用等内容。 13．计量经济学模型用于预测前必须通过的检验分别是__________检验、__________检验、__________检验和__________检验。 14．计量经济模型的计量经济检验通常包括随机误差项的__________检验、__________检验、解释变量的__________检验。 15．计量经济学模型的应用可以概括为四个方面，即__________、__________、__________、__________。 16．结构分析所采用的主要方法是__________、__________和__________。 二、单选题： 1．计量经济学是一门（）学科。 A.数学 B.经济 C.统计 D.测量

计量经济学题库(超完整版)及答案【强力修正版】

计量经济学题库 一、单项选择题（每小题1分） 1．计量经济学是下列哪门学科的分支学科（）。 A．统计学 B．数学 C．经济学 D．数理统计学 2．计量经济学成为一门独立学科的标志是（）。 A．1930年世界计量经济学会成立B．1933年《计量经济学》会刊出版 C．1969年诺贝尔经济学奖设立 D．1926年计量经济学（Economics）一词构造出来 3．外生变量和滞后变量统称为（）。 A．控制变量 B．解释变量 C．被解释变量 D．前定变量 4．横截面数据是指（）。 A．同一时点上不同统计单位相同统计指标组成的数据B．同一时点上相同统计单位相同统计指标组成的数据 C．同一时点上相同统计单位不同统计指标组成的数据D．同一时点上不同统计单位不同统计指标组成的数据 5．同一统计指标，同一统计单位按时间顺序记录形成的数据列是（）。 A．时期数据 B．混合数据 C．时间序列数据 D．横截面数据 6．在计量经济模型中，由模型系统内部因素决定，表现为具有一定的概率分布的随机变量，其数值受模型中其他变量影响的变量是（）。 A．内生变量 B．外生变量 C．滞后变量 D．前定变量 7．描述微观主体经济活动中的变量关系的计量经济模型是（）。 A．微观计量经济模型 B．宏观计量经济模型 C．理论计量经济模型 D．应用计量经济模型 8．经济计量模型的被解释变量一定是（）。 A．控制变量 B．政策变量 C．内生变量 D．外生变量 9．下面属于横截面数据的是（）。 A．1991－2003年各年某地区20个乡镇企业的平均工业产值 B．1991－2003年各年某地区20个乡镇企业各镇的工业产值 C．某年某地区20个乡镇工业产值的合计数 D．某年某地区20个乡镇各镇的工业产值 10．经济计量分析工作的基本步骤是（）。 A．设定理论模型→收集样本资料→估计模型参数→检验模型B．设定模型→估计参数→检验模型→应用模型 C．个体设计→总体估计→估计模型→应用模型D．确定模型导向→确定变量及方程式→估计模型→应用模型 11．将内生变量的前期值作解释变量，这样的变量称为（）。 A．虚拟变量 B．控制变量 C．政策变量 D．滞后变量 12．（）是具有一定概率分布的随机变量，它的数值由模型本身决定。 A．外生变量 B．内生变量 C．前定变量 D．滞后变量 13．同一统计指标按时间顺序记录的数据列称为（）。 A．横截面数据 B．时间序列数据 C．修匀数据 D．原始数据 14．计量经济模型的基本应用领域有（）。 A．结构分析、经济预测、政策评价 B．弹性分析、乘数分析、政策模拟 C．消费需求分析、生产技术分析、 D．季度分析、年度分析、中长期分析 15．变量之间的关系可以分为两大类，它们是（）。 A．函数关系与相关关系B．线性相关关系和非线性相关关系 C．正相关关系和负相关关系D．简单相关关系和复杂相关关系 16．相关关系是指（）。 A．变量间的非独立关系B．变量间的因果关系C．变量间的函数关系 D．变量间不确定性的依存关系17．进行相关分析时的两个变量（）。

计量经济学答案

一、名词解释 1.时间序列数据的平稳性:如果随机时间序列均值和方差均是与时间t无关的常数，协方差只与时间间隔k有关，则称该随机时间序列是平稳的。 2.虚拟变量：是指人们构造的反应定性因素变化、只取0和1的人工变量，并且习惯上用符号D来表示。 3.异方差性：对于不同的样本点，随机误差项的方差不等于常数，则称模型出现了异方差性。 4.自相关性：如果随机误差项的各期值之间存在着相关关系，即协方差不等于0，则称模型存在着自相关性。 5随机变量的协整关系：如果同阶单整序列线性组合后单整阶数降低，则称变量之间存在着协整关系。 6.给定一个信息集，At，它至少包含（Xt，Yt），在“现在和过去可以影响未来，而未来不能影响过去”城里下，如果利用Xt的过去比不利用它时可以更好地预测Yt，称Xt为Yt的格兰杰原因，反之亦然。 7.随机变量的协整性： 8. 条件异方差ARCH模型：考虑m阶自回归模型AR（m） Yt=c+ρ1yt-1+ρ2yt-2+……+ρmyt-m+εt 其中εt为白噪声过程 随机误差项的平方（εt）2服从一个q阶自回归过程，即 （εt）2=α0＋α1（εt-1）2+α2（εt-2）2+……+αq（εt-p）2+ηt （1） 其中ηt服从白噪声过程。对模型的一个约束条件是（1）的特征方程 1-α1z-α2z2-……-αq Z q=0 的所有根均落在单位圆外，即要求模型参数满足 其中α1+α2+……αq＜1 此外，为保证εt2为正值，对模型的另一个约束条件为α0＞0，αi≥0，1≤i≤q。上述模型即为条件方差模型。 9.误差修正模型ECM: 对于yi的（1，1）阶自回归滞后模型： εi Y t=α+β0x t+β1x t-1+β2y t-1+ ⊿y =β0⊿x t+γecm t-1+εt 。(1) 其中，ecm t-1=y t-1-α0-α1x t-1 ，γ=β2-1，α0=（α+ t β0）/﹙1-β2﹚，α1=β1/（1-β2） 称式（1）为误差修正模型ECM 10.多重共线性:多元回归模型的解释变量之间存在较强的线性关系的性质 二、填空题 1.合理选择解释变量的关键：正确理解有关经济理论和把握所研究经济现象的行为规律。 2.计量经济模型的用途一般包括：结构分析、经济预测、政策评价、实证分析。 3.计量经济模型检验的内容一般包括：经济检验、统计检验、计量经济检验、预测性能检验。 4.对于不可直接线性化的非线性模型的处理方法： 对于可间接线性化的模型，可以通过Cobb-Douglas生产函数模型、Logistic模型变换成标准的线性模型；对于不可线性化的模型，可以通过Toylor技术展开法、非线性最小二乘法来求得参数估计值。

计量经济学习题及参考答案解析详细版

计量经济学（第四版）习题参考答案 潘省初

第一章 绪论 试列出计量经济分析的主要步骤。 一般说来，计量经济分析按照以下步骤进行： （1）陈述理论（或假说） （2）建立计量经济模型 （3）收集数据 （4）估计参数 （5）假设检验 （6）预测和政策分析 计量经济模型中为何要包括扰动项? 为了使模型更现实，我们有必要在模型中引进扰动项u 来代表所有影响因变量的其它因素，这些因素包括相对而言不重要因而未被引入模型的变量，以及纯粹的随机因素。 什么是时间序列和横截面数据? 试举例说明二者的区别。 时间序列数据是按时间周期（即按固定的时间间隔）收集的数据，如年度或季度的国民生产总值、就业、货币供给、财政赤字或某人一生中每年的收入都是时间序列的例子。 横截面数据是在同一时点收集的不同个体（如个人、公司、国家等）的数据。如人口普查数据、世界各国2000年国民生产总值、全班学生计量经济学成绩等都是横截面数据的例子。 估计量和估计值有何区别？ 估计量是指一个公式或方法，它告诉人们怎样用手中样本所提供的信息去估计总体参数。在一项应用中，依据估计量算出的一个具体的数值，称为估计值。如Y 就是一个估计量，1 n i i Y Y n == ∑。现有一样本，共4个数，100，104，96，130，则 根据这个样本的数据运用均值估计量得出的均值估计值为 5.1074 130 96104100=+++。 第二章 计量经济分析的统计学基础 略，参考教材。

请用例中的数据求北京男生平均身高的99％置信区间 N S S x = = 4 5= 用 =，N-1=15个自由度查表得005.0t =，故99%置信限为 x S t X 005.0± =174±×=174± 也就是说，根据样本，我们有99%的把握说，北京男高中生的平均身高在至厘米之间。 25个雇员的随机样本的平均周薪为130元，试问此样本是否取自一个均值为120元、标准差为10元的正态总体？ 原假设 120:0=μH 备择假设 120:1≠μH 检验统计量 () 10/2510/25 X X μσ-Z == == 查表96.1025.0=Z 因为Z= 5 >96.1025.0=Z ,故拒绝原假设, 即 此样本不是取自一个均值为120元、标准差为10元的正态总体。 某月对零售商店的调查结果表明，市郊食品店的月平均销售额为2500元，在下一个月份中，取出16个这种食品店的一个样本，其月平均销售额为2600元，销售额的标准差为480元。试问能否得出结论，从上次调查以来，平均月销售额已经发生了变化？ 原假设 : 2500:0=μH 备择假设 : 2500:1≠μH ()100/1200.83?480/16 X X t μσ-= === 查表得 131.2)116(025.0=-t 因为t = < 131.2=c t , 故接受原假 设,即从上次调查以来，平均月销售额没有发生变化。

计量经济学第二章主要公式

第二章主要公式 资料地址：https://www.sodocs.net/doc/663339276.html,/jl 1、回归模型概述 （1）相关分析与回归分析 经济变量之间的关系：函数关系、相关关系 相关关系：单相关和复相关，完全相关、不完全相关和不相关，正相关与负相关，线性相关和负相关，线性相关和非线性相关。 相关分析： ——总体相关系数XY ρ= ——样本相关系数()() n i i XY X X Y Y r --= ∑ ——多个变量之间的相关程度可用复相关系数和偏相关系数度量 回归分析：相关关系 + 因果关系 （2）随机误差项：含有随机误差项是计量经济学模型与数理经济学模型的一大区别。 （3）总体回归模型 总体回归曲线：给定解释变量条件下被解释变量的期望轨迹。 总体回归函数：(|)()i i E Y X f X = 总体回归模型：(|)()i i i i i Y E Y X f X μμ=+=+ 线性总体回归模型：011,2,...,i i i Y X i n ββμ=++= （4）样本回归模型 样本回归曲线：根据样本回归函数得到的被解释变量的轨迹。 （线性）样本回归函数： 01???i i Y X ββ=+ （线性）样本回归模型：01???i i i Y X e ββ=++ 2、一元线性回归模型的参数估计 （1）基本假设 ① 解释变量：是确定性变量，不是随机变量 var()0i X = ② 随机误差项：零均值、同方差，在不同样本点之间独立，不存在序列相关等 ()01,2,...,i E i n μ== 2var()1,2,...,i i n μσ==

cov(,)0;,1,2,...,i j i j i j n μμ=≠= ③ 随机误差项与解释变量：不相关 cov(,)01,2,...,i i X i n μ== ④ （针对最大似然法和假设检验）随机误差项： 2~(0,)1,2,...,i N i n μσ= ⑤ 回归模型正确设定。 【前四条为线性回归模型的古典假设，即高斯假设。满足古典假设的线性回归模型称为古典线性回归模型。】 （2）参数的普通最小二乘估计（OLS ） 目标：21 min n i i e =∑ 对于一元线性回归模型：011,2,...,i i i Y X i n ββμ=++= 正规方程组： 011 011 ?? 2[()]0??2[()]0n i i i n i i i i Y X X Y X ββββ==?--+=????--+=??∑∑ 解得： 011 112 211??()()?()n n i i i i i i n n i i i i Y X X X Y Y x y X X x βββ====?=-???--?==??-?? ∑∑∑∑ （3）最大似然估计（ML ） 对于一元线性回归模型：011,2,...,i i i Y X i n ββμ=++= 重要的基本假设： 2~(0,)1,2,...,cov(,)0;,1,2,...,var()01,2,...,i i j i N i n i j i j n X i n μσμμ?=? =≠=?? ==? 得到：2 01~(,)1,2,...,i i Y N X i n ββσ+= 【且cov(,)0;,1,2,...,i j Y Y i j i j n =≠=，这个对最大似然法的估计很重要】 则目标：12,,...,n Y Y Y 的联合概率密度最大，即

计量经济学题库及答案

四、简答题（每小题5分） 1．简述计量经济学与经济学、统计学、数理统计学学科间的关系。2．计量经济模型有哪些应用？ 3．简述建立与应用计量经济模型的主要步骤。 4．对计量经济模型的检验应从几个方面入手？ 5．计量经济学应用的数据是怎样进行分类的？ 6．在计量经济模型中，为什么会存在随机误差项？ 7．古典线性回归模型的基本假定是什么？ 8．总体回归模型与样本回归模型的区别与联系。 9．试述回归分析与相关分析的联系和区别。 10．在满足古典假定条件下，一元线性回归模型的普通最小二乘估计量有哪些统计性质？ 11．简述BLUE 的含义。 12．对于多元线性回归模型，为什么在进行了总体显著性F 检验之后，还要对每个回归系数进行是否为0的t 检验？ 13.给定二元回归模型：，请叙述模型的古典假定。 14.在多元线性回归分析中，为什么用修正的决定系数衡量估计模型对样本观测值的拟合优度？ 15.修正的决定系数2R 及其作用。 16.常见的非线性回归模型有几种情况？ 17.观察下列方程并判断其变量是否呈线性，系数是否呈线性，或都是或都不是。 ①t t t u x b b y ++=310 ②t t t u x b b y ++=log 10 ③ t t t u x b b y ++=log log 10 ④t t t u x b b y +=)/(10 18. 观察下列方程并判断其变量是否呈线性，系数是否呈线性，或都是或都不是。 ①t t t u x b b y ++=log 10 ②t t t u x b b b y ++=)(210 ③ t t t u x b b y +=)/(10 ④t b t t u x b y +-+=)1(110 19.什么是异方差性？试举例说明经济现象中的异方差性。 20.产生异方差性的原因及异方差性对模型的OLS 估计有何影响。 21.检验异方差性的方法有哪些？ 22.异方差性的解决方法有哪些？ 23.什么是加权最小二乘法？它的基本思想是什么？ 24.样本分段法（即戈德菲尔特——匡特检验）检验异方差性的基本原理及其使用条件。 25．简述DW 检验的局限性。 26．序列相关性的后果。 27．简述序列相关性的几种检验方法。

伍德里奇计量经济学第六版答案Appendix-E

271 APPENDIX E SOLUTIONS TO PROBLEMS E.1 This follows directly from partitioned matrix multiplication in Appendix D. Write X = 12n ?? ? ? ? ? ???x x x , X ' = (1'x 2'x n 'x ), and y = 12n ?? ? ? ? ? ??? y y y Therefore, X 'X = 1 n t t t ='∑x x and X 'y = 1 n t t t ='∑x y . An equivalent expression for ?β is ?β = 1 11n t t t n --=??' ???∑x x 11n t t t n y -=??' ??? ∑x which, when we plug in y t = x t β + u t for each t and do some algebra, can be written as ?β= β + 1 11n t t t n --=??' ???∑x x 11n t t t n u -=??' ??? ∑x . As shown in Section E.4, this expression is the basis for the asymptotic analysis of OLS using matrices. E.2 (i) Following the hint, we have SSR(b ) = (y – Xb )'(y – Xb ) = [?u + X (?β – b )]'[ ?u + X (?β – b )] = ?u '?u + ?u 'X (?β – b ) + (?β – b )'X '?u + (?β – b )'X 'X (?β – b ). But by the first order conditions for OLS, X '?u = 0, and so (X '?u )' = ?u 'X = 0. But then SSR(b ) = ?u '?u + (?β – b )'X 'X (?β – b ), which is what we wanted to show. (ii) If X has a rank k then X 'X is positive definite, which implies that (?β – b ) 'X 'X (?β – b ) > 0 for all b ≠ ?β . The term ?u '?u does not depend on b , and so SSR(b ) – SSR(?β) = (?β– b ) 'X 'X (?β – b ) > 0 for b ≠?β. E.3 (i) We use the placeholder feature of the OLS formulas. By definition, β = (Z 'Z )-1Z 'y = [(XA )' (XA )]-1(XA )'y = [A '(X 'X )A ]-1A 'X 'y = A -1(X 'X )-1(A ')-1A 'X 'y = A -1(X 'X )-1X 'y = A -1?β . (ii) By definition of the fitted values, ?t y = ?t x β and t y = t z β. Plugging z t and β into the second equation gives t y = (x t A )(A -1?β ) = ?t x β = ?t y . (iii) The estimated variance matrix from the regression of y and Z is 2σ(Z 'Z )-1 where 2σ is the error variance estimate from this regression. From part (ii), the fitted values from the two

计量经济学答案部分Word版

第一章导论 一、单项选择题 1-6: CCCBCAC 二、多项选择题 ABCD;ACD;ABCD 三.问答题 什么是计量经济学？ 答案见教材第3页 四、案例分析题 假定让你对中国家庭用汽车市场发展情况进行研究，应该分哪些步骤，分别如何分析？（参考计量经济学研究的步骤） 第一步：选取被研究对象的变量：汽车销售量 第二步：根据理论及经验分析，寻找影响汽车销售量的因素，如汽车价格，汽油价格，收入水平等 第三步：建立反映汽车销售量及其影响因素的计量经济学模型 第四步：估计模型中的参数； 第五步：对模型进行计量经济学检验、统计检验以及经济意义检验； 第六步：进行结构分析及在给定解释变量的情况下预测中国汽车销售量的未来值为汽车业的发展提供政策实施依据。 第二章简单线性回归模型 一、填空题 1、线性、无偏、最小方差性（有效性），BLUE。 2、解释变量；参数；参数。 3、随机误差项；随机误差项。 二、单项选择题 1-4：BBDA;6-11：CDCBCA 三、多项选择题 1.ABC; 2.ABC; 3.BC; 4.ABE; 5.AD; 6.BC 四、判断正误： 1. 错； 2. 错； 3. 对； 4.错； 5. 错； 6. 对； 7. 对； 8.错 五、简答题： 1.为什么模型中要引入随机扰动项？ 答：模型是对经济问题的一种数学模型，在模型中，被解释变量是研究的对象，解释变量是其确定的解释因素，但由于实际问题的错综复杂，影响被解释变量的因素中，除了包括在模型中的解释变量以外，还有其他一些因素未能包括在模型中，但却影响被解释变量，我们把这类变量统一用随机误差项表示。随机误差项包含的因素有：

伍德里奇---计量经济学第8章部分计算机习题详解(STATA)

班级：金融学×××班姓名：××学号：×××××××C8.1SLEEP75.RAW sleep=β0+β1totwork+β2educ+β3age+β4age2+β5yngkid+β6male+u 解：（ⅰ）写出一个模型，容许u的方差在男女之间有所不同。这个方差不应该取决于其他因素。 在sleep=β0+β1totwork+β2educ+β3age+β4age2+β5yngkid+β6male+u模型下，u方差要取决于性别，则可以写成：Var u︳totwork,educ,age,yngkid,male =Var u︳male =δ0+δ1male。所以，当方差在male=1时，即为男性时，结果为δ0+δ1；当为女性时，结果为δ0。 将sleep对totwork，educ，age，age2，yngkid和male进行回归，回归结果如下： （ⅱ）利用SLEEP75.RAW的数据估计异方差模型中的参数。u的估计方差对于男人和女人而言哪个更高？ 由截图可知：u2=189359.2?28849.63male+r

20546.36 (27296.36) 由于male 的系数为负，所以u 的估计方差对女性而言更大。 （ⅲ）u 的方差是否对男女而言有显著不同？ 因为male 的 t 统计量为?1.06，所以统计不显著，故u 的方差是否对男女而言并没有显著不同。 C8.2 HPRICE1.RAW price =β0+β1lotsize +β2sqrft +β3bdrms +u 解：（ⅰ）利用HPRICE 1.RAW 中的数据得到方程（8.17）的异方差—稳健的标准误。讨论其与通常的标准误之间是否存在任何重要差异。 ● 先进行一般回归，结果如下： ● 再进行稳健回归，结果如下： 由两个截图可得：price =?21.77+0.00207lotsize +0.123sqrft +13.85bdrms 29.48 0.00064 0.013 (9.01) 37.13 0.00122 0.018 [8.48] n = 88， R 2=0.672 比较稳健标准误和通常标准误，发现lotsize 的稳健标准误是通常下的2倍，使得 t 统计量相差较大。而sqrft 的稳健标准误也比通常的大，但相差不大，bdrms 的稳健标准误比通常的要小些。 （ⅱ）对方程（8.18）重复第（ⅰ）步操作。 n =706，R 2=0.0016

计量经济学-案例分析-第二章

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计学(第六版)第七章课后练习答案

第七章 课后练习答案 7.1 （1）已知：96.1%,951,25,40,52/05.0==-===z x n ασ。 样本均值的抽样标准差79.0405== = n x σ σ （2）边际误差55.140 5 96.12/=? ==n z E σ α 7.2 （1）已知：96.1%,951,120,49,152/05.0==-===z x n ασ。 样本均值的抽样标准差14.249 15== = n x σ σ （2）边际误差20.449 1596.12 /=? ==n z E σ α （3）由于总体标准差已知，所以总体均值μ的95%的置信区间为 20.412049 1596.11202 /±=? ±=±n z x σ α 即()2.124,8.115 7.3 已知：96.1%,951,104560,100,854142/05.0==-===z x n ασ。 由于总体标准差已知，所以总体均值μ的95%的置信区间为 144.16741104560100 8541496.11045602 /±=? ±=±n z x σ α 即)144.121301,856.87818( 7.4 （1）已知：645.1%,901,12,81,1002/1.0==-===z s x n α。 由于100=n 为大样本，所以总体均值μ的90%的置信区间为： 974.181100 12645.1812 /±=? ±=±n s z x α 即)974.82,026.79(

（2）已知：96.1%,951,12,81,1002/05.0==-===z s x n α。 由于100=n 为大样本，所以总体均值μ的95%的置信区间为： 352.281100 1296.1812 /±=? ±=±n s z x α 即)352.83,648.78( （3）已知：58.2%,991,12,81,1002/05.0==-===z s x n α。 由于100=n 为大样本，所以总体均值μ的99%的置信区间为： 096.381100 1258.2812 /±=? ±=±n s z x α 即)096.84,940.77( 7.5 （1）已知：96.1%,951,5.3,25,602/05.0==-===z x n ασ。 由于总体标准差已知，所以总体均值μ的95%的置信区间为： 89.02560 5.39 6.1252 /±=? ±=±n z x σ α 即)89.25,11.24( （2）已知：33.2%,981,89.23,6.119,752/02.0==-===z s x n α。 由于75=n 为大样本，所以总体均值μ的98%的置信区间为： 43.66.11975 89.2333.26.1192 /±=? ±=±n s z x α 即)03.126,17.113( （3）已知：645.1%,901,974.0,419.3,322/1.0==-===z s x n α。 由于32=n 为大样本，所以总体均值μ的90%的置信区间为： 283.0419.332 974.0645.1419.32 /±=? ±=±n s z x α 即)702.3,136.3(

计量经济学教程(赵卫亚)课后答案第二章汇编

第二章 回归模型思考与练习参考答案 2.1参考答案 ⑴答：解释变量为确定型变量、互不相关（无多重共线性）；随机误差项零的值、同方差、非自相关；解释变量与随机误差项不相关。 现实经济中，这些假定难以成立。要解决这些问题就得对古典回归理论做进一步发展，这就产生了现代回归理论。 ⑵答：总体方差是总体回归模型中随机误差项i ε的方差；参数估计误差则属于样本回归模型中的概念，通常是指参数估计的均方误。参数估计的均方误为 MSE ()i i b b ?=E ()2?i i b b -=D ()i b ?=()[]ii u 12-'χχσ 即根据参数估计的无偏线，参数估计的均方误与其方差相等。而参数估计的方差又源于总体方差。因此，参数估计误差是总体方差的表现，总体方差是参数估计误差的根源。 ⑶答：总体回归模型 ()i i i x y E y ε+= 样本回归模型i i i e y y +=? i ε是因变量y 的个别值i y 与因变量y 对i x 的总体回归函数值() i x y E 的偏差；i e 为因变量y 的观测值i y 与因变量y 的样本回归函数值i y ?的偏差。 i e 在概念上类似于i ε，是对i ε的估计。 对于既定理论模型，OLS 法能使模型估计的拟和误差达最小。但或许我们可选择更理想的理论模型，从而进一步提高模型对数据的拟和程度。 ⑷答：2R 检验说明模型对样本数据的拟和程度；F 检验说明模型对总体经济关系的近似程度。 ()()()k k n R R k n Model Total k Model k m Error k Model F 111122--?-=---=--= 由02>??R F 可知，F 是2R 的单调增函数。对每一个临界值?F ，都可以找到一个2?R 与之对应，当22?>R R 时便有?>F F 。 ⑸答：在古典回归模型假定成立的条件下，OLS 估计是所有的线形无偏估计量中的有效估计量。 ⑹答：如果模型通过了F 检验，则表明模型中所有解释变量对被解释变量的影响显著。但这并不说明多个解释变量的影响都是显著的。建模开始时，常根据先验知识尽可能找出影响被解释变量的所有因素，这样就可能会选择不重要的因素作为解释变量。对单个解释变量的显著性检验可以剔除这些不重要的影响因素。 ⑺答：考虑两个经济变量y 与x ，及一组观测值(){},,2,1,,n i y x i i =。

计量经济学导论：现代观点第四版习题答案

DATA SET HANDBOOK Introductory Econometrics: A Modern Approach, 4e Jeffrey M. Wooldridge This document contains a listing of all data sets that are provided with the fourth edition of Introductory Econometrics: A Modern Approach. For each data set, I list its source (wherever possible), where it is used or mentioned in the text (if it is), and, in some cases, notes on how an instructor might use the data set to generate new homework exercises, exam problems, or term projects. In some cases, I suggest ways to improve the data sets. Special thanks to Edmund Wooldridge, who provided valuable assistance in updating the page numbers for the fourth edition. 401K.RAW Source:L.E. Papke (1995), “Participation in and Contributions to 401(k) Pension Plans: Evidence from Plan Data,”Journal of Human Resources 30, 311-325. Professor Papke kindly provided these data. She gathered them from the Internal Revenue Service’s Form 5500 tapes. Used in Text: pages 64, 80, 135-136, 173, 217, 685-686 Notes: This data set is used in a variety of ways in the text. One additional possibility is to investigate whether the coefficients from the regression of prate on mrate, log(totemp) differ by whether the plan is a sole plan. The Chow test (see Section 7.4), and the less restrictive version that allows different intercepts, can be used. 401KSUBS.RAW Source: A. Abadie (2003), “Semiparametric Instrumental Variable Estimation of Treatment Response Models,”Journal of Econometrics 113, 231-263. Professor Abadie kindly provided these data. He obtained them from the 1991 Survey of Income and Program Participation (SIPP). Used in Text: pages 165, 182, 222, 261, 279-280, 288, 298-299, 336, 542 Notes: This data set can also be used to illustrate the binary response models, probit and logit, in Chapter 17, where, say, pira (an indicator for having an individual retirement account) is the dependent variable, and e401k [the 401(k) eligibility indicator] is the key explanatory variable.

第二章习题及答案-计量经济学

第二章 简单线性回归模型 一、单项选择题（每题2分）： 1、回归分析中定义的（ ）。 A 、解释变量和被解释变量都是随机变量 B 、解释变量为非随机变量，被解释变量为随机变量 C 、解释变量和被解释变量都为非随机变量 D 、解释变量为随机变量，被解释变量为非随机变量 2、最小二乘准则是指使（ ）达到最小值的原则确定样本回归方程。 A 、1 ?()n t t t Y Y =-∑ B 、1?n t t t Y Y = -∑ C 、?max t t Y Y - D 、21 ?()n t t t Y Y =-∑ 3、下图中“{”所指的距离是（ ）。 A 、随机误差项 B 、残差 C 、i Y 的离差 D 、?i Y 的离差 4、参数估计量?β是i Y 的线性函数称为参数估计量具有( )的性质。 A 、线性 B 、无偏性 C 、有效性 D 、一致性 5、参数β的估计量β? 具备最佳性是指（ ）。 A 、0)?(=βVar B 、)? (βVar 为最小 C 、0?=-ββ D 、)? (ββ-为最小 6、反映由模型中解释变量所解释的那部分离差大小的是( )。 A 、总体平方和 B 、回归平方和 C 、残差平方和 D 、样本平方和 7、总体平方和TSS 、残差平方和RSS 与回归平方和ESS 三者的关系是（ ）。 X 1?β+ i Y

A 、RSS=TSS+ESS B 、TSS=RSS+ESS C 、ESS=RSS-TSS D 、ESS=TSS+RSS 8、下面哪一个必定是错误的（ ）。 A 、 i i X Y 2.030? += ，8.0=XY r B 、 i i X Y 5.175?+-= ，91.0=XY r C 、 i i X Y 1.25? -=，78.0=XY r D 、 i i X Y 5.312?--=，96.0-=XY r 9、产量（X ，台）与单位产品成本（Y ，元/台）之间的回归方程为?356 1.5Y X =-，这说明（ ）。 A 、产量每增加一台，单位产品成本增加356元 B 、产量每增加一台，单位产品成本减少1.5元 C 、产量每增加一台，单位产品成本平均增加356元 D 、产量每增加一台，单位产品成本平均减少1.5元 10、回归模型i i i X Y μββ++=10，i = 1，…，n 中，总体方差未知，检验 010=β：H 时，所用的检验统计量1 ? 1 1?βββS -服从（ ）。 A 、）（22 -n χ B 、）（1-n t C 、）（12-n χ D 、）（2-n t 11、对下列模型进行经济意义检验，哪一个模型通常被认为没有实际价值的（ ）。 A 、i C （消费）i I 8.0500+=（收入） B 、di Q （商品需求）i I 8.010+=（收入）i P 9.0+（价格） C 、si Q （商品供给）i P 75.020+=（价格） D 、i Y （产出量）6.065.0i K =（资本）4 .0i L （劳动） 12、进行相关分析时，假定相关的两个变量( )。 A 、都是随机变量 B 、都不是随机变量 C 、一个是随机变量，一个不是随机变量 D 、随机或非随机都可以 13、假设用OLS 法得到的样本回归直线为i i i e X Y ++=2 1 ??ββ ，以下说法不正确的是( )。 A 、∑=0i e B 、),(Y X 一定在回归直线上 C 、Y Y =? D 、0),(≠i i e X COV 14、对样本的相关系数γ，以下结论错误的是（ ）。 A 、γ越接近0，X 和Y 之间的线性相关程度越高

安徽财经大学计量经济学 第二章练习题及参考解答

第二章练习题及参考解答 2.1 为研究中国的货币供应量(以货币与准货币M2表示)与国内生产总值(GDP)的相互依存关系，分析表中1990年—2007年中国货币供应量（M2）和国内生产总值（GDP ）的有关数据: 表2.9 1990年—2007年中国货币供应量和国内生产总值(单位:亿元) 资料来源:中国统计年鉴2008,中国统计出版社 对货币供应量与国内生产总值作相关分析，并说明相关分析结果的经济意义。 练习题2.1 参考解答: 计算中国货币供应量(以货币与准货币M2表示)与国内生产总值(GDP)的相关系数为: 计算方法: XY n X Y X Y r -= 或 ,()()X Y X X Y Y r --= 计算结果: M2 GDP M2 1 0.996426148646 GDP 0.996426148646 1 经济意义: 这说明中国货币供应量与国内生产总值(GDP)的线性相关系数为0.996426,线性

相关程度相当高。 2.2 为研究美国软饮料公司的广告费用X与销售数量Y的关系,分析七种主要品牌软饮料公司的有关数据 表2.10 美国软饮料公司广告费用与销售数量 资料来源:(美) Anderson D R等. 商务与经济统计.机械工业出版社.1998. 405 绘制美国软饮料公司广告费用与销售数量的相关图, 并计算相关系数，分析其相关程度。能否在此基础上建立回归模型作回归分析？ 练习题2.2参考解答 美国软饮料公司的广告费用X与销售数量Y的散点图为 说明美国软饮料公司的广告费用X与销售数量Y正线性相关。

若以销售数量Y 为被解释变量,以广告费用X 为解释变量,可建立线性回归模型 i i i u X Y ++=21ββ 利用EViews 估计其参数结果为 x 4036.147857.21y ?+= （96.9800）（1.3692） t= (-0.131765) (10.5200) 9568.02=R F=110.6699 S.E=92302.73 D.W=1.4389 经t 检验表明, 广告费用X 对美国软饮料公司的销售数量Y 确有显著影响。回归结果表明，广告费用X 每增加1百万美元, 平均说来软饮料公司的销售数量将增加14.40359(百万箱)。 2.3 为了研究深圳市地方预算内财政收入与国内生产总值的关系，得到以下数据： 表2.11 深圳市地方预算内财政收入与国内生产总值